Question

How do you implicitly differentiate `-y= x^3y^2-3x^2y^2+xy^4`?

Answer 1

`dy/dx = (6xy^2 -y^4 - 3x^2y^2)/(1 + 2x^3y - 6x^2y + 4xy^3)`

Explanation:

`-y = x^3y^2 - 3x^2y^2 + xy^4`

We differentiate everything wrt `x`:

`-d/dx(y) = d/dx(x^3y^2) - d/dx(3x^2y^2) + d/dx(xy^4)`

We can just deal with the first term;

`-dy/dx = d/dx(x^3y^2) - d/dx(3x^2y^2) + d/dx(xy^4)`

For the other term we apply the product rule;

`-dy/dx = {(x^3)(d/dxy^2) + (d/dxx^3)(y^2)} - {(3x^2)(d/dxy^2) + (d/dx3x^2)(y^2)} + {(x)(d/dxy^4) + (d/dxx)(y^4)}`

`:. -dy/dx = x^3(d/dxy^2) + 3x^2y^2 - 3x^2(d/dxy^2) + 6xy^2 + x(d/dxy^4) + y^4`

Next we use the chain rule so that we can differentiate wrt `y` (this is the "Implicit" part of the differentiation
`:. -dy/dx = x^3(dy/dxd/dyy^2) + 3x^2y^2 - 3x^2(dy/dxd/dyy^2) - 6xy^2 + x(dy/dxd/dyy^4) + y^4`

And now we can differentiate the `y` terms wrt `y`
`:. -dy/dx = x^3(dy/dx2y) + 3x^2y^2 - 3x^2(dy/dx2y) - 6xy^2 + x(dy/dx4y^3) + y^4`

Finally, we can tidy up, collect terms and factorise `dy/dx` to form an explicit solution

`:. -dy/dx = 2x^3ydy/dx + 3x^2y^2 - 6x^2ydy/dx - 6xy^2 + 4xy^3dy/dx + y^4`

`:. dy/dx + 2x^3ydy/dx - 6x^2ydy/dx + 4xy^3dy/dx = 6xy^2 -y^4 - 3x^2y^2`

`:. dy/dx(1 + 2x^3y - 6x^2y + 4xy^3) = 6xy^2 -y^4 - 3x^2y^2`
`:. dy/dx = (6xy^2 -y^4 - 3x^2y^2)/(1 + 2x^3y - 6x^2y + 4xy^3)`